Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $n = \dfrac{x^2 - 8x + 7}{x^2 - 7x} \times \dfrac{x - 5}{2x - 2} $
Solution: First factor the quadratic. $n = \dfrac{(x - 1)(x - 7)}{x^2 - 7x} \times \dfrac{x - 5}{2x - 2} $ Then factor out any other terms. $n = \dfrac{(x - 1)(x - 7)}{x(x - 7)} \times \dfrac{x - 5}{2(x - 1)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (x - 1)(x - 7) \times (x - 5) } { x(x - 7) \times 2(x - 1) } $ $n = \dfrac{ (x - 1)(x - 7)(x - 5)}{ 2x(x - 7)(x - 1)} $ Notice that $(x - 7)$ and $(x - 1)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ \cancel{(x - 1)}(x - 7)(x - 5)}{ 2x(x - 7)\cancel{(x - 1)}} $ We are dividing by $x - 1$ , so $x - 1 \neq 0$ Therefore, $x \neq 1$ $n = \dfrac{ \cancel{(x - 1)}\cancel{(x - 7)}(x - 5)}{ 2x\cancel{(x - 7)}\cancel{(x - 1)}} $ We are dividing by $x - 7$ , so $x - 7 \neq 0$ Therefore, $x \neq 7$ $n = \dfrac{x - 5}{2x} ; \space x \neq 1 ; \space x \neq 7 $